Problem: Given that $\cos x - 4 \sin x = 1,$ find all possible values of $\sin x + 4 \cos x.$
Solution: From the equation $\cos x - 4 \sin x = 1,$
\[\cos x - 1 = 4 \sin x.\]Squaring both sides, we get
\[\cos^2 x - 2 \cos x + 1 = 16 \sin^2 x = 16 - 16 \cos^2 x.\]This simplifies to $17 \cos^2 x - 2 \cos x - 15 = 0,$ which factors as
\[(\cos x - 1)(17 \cos x + 15) = 0.\]Hence, $\cos x = 1$ or $\cos x = -\frac{15}{17}.$

If $\cos x = 1,$ then $\sin x = \frac{\cos x - 1}{4} = 0,$ so
\[\sin x + 4 \cos x = 0 + 4(1) = \boxed{4}.\]If $\cos x = -\frac{15}{17},$ then $\sin x = \frac{\cos x - 1}{4} = -\frac{8}{17},$ so
\[\sin x + 4 \cos x = -\frac{8}{17} + 4 \left( -\frac{15}{17} \right) = \boxed{-4}.\]